### Video Transcript

Given that vector π¨ equals
negative three π’ minus five π£ plus π€, vector π© equals negative five π’ minus
three π£ minus three π€, vector π equals negative two π’ minus π£ plus four π€, and
π¨ plus π times π© is perpendicular to the vector π, determine π.

Given these three three-dimensional
vectors π¨, π©, and π, we know that if we take vector π¨ and we add it to this
multiple π of vector π©, then that result is perpendicular to vector π. Whenever two vectors are
perpendicular to one another, it means something very specific. Because the two vectors π¨ plus π
times π© and π are perpendicular, we can say that the dot product between them
equals zero.

In this expression, weβre given all
the values for our three vectors. But we donβt yet know the value of
this multiple π. Thatβs what we want to solve
for. And we can start to do that by
substituting in the components of vectors π¨, π©, and π. When we do so, we get this big long
expression. The part weβll focus on for now is
where we have vector π¨ being added to this multiple π of vector π©.

As a first step, letβs multiply the
components of π© by this factor π. Next, we can add together these two
vectors, adding them component by component. For the π’-component, we have
negative three minus five π; the π£-component is negative five minus three π; and
then for the π€-component positive one minus three π.

Weβre now ready to go ahead and
compute this dot product. We do this by multiplying together
the π’-components, then adding that to the multiple of the π£-components, and adding
that to the multiple of the π€-components. That gives us this expression. And working on the left-hand side,
we can combine together six, five, and four β thatβs 15 β and 10π, three π, and
negative 12π β thatβs simply π. And as weβve seen, this is equal to
zero. That tells us that π equals
negative 15.

So, given these vectors π¨, π©, and
π, for π¨ plus π times π© to be perpendicular to π, π equals negative 15.